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Defective items in a production process are items that don't meet the required quality standards. In the given problem, **defective screws** are screws that don't perform as needed. When dealing with probabilities of defects, we need to focus on the likelihood of picking a defective item from a batch. Here, the probability of any single screw being defective is given as **0.2**. This means if you picked one screw at random, there is a 20% chance it would be defective.

By determining the number of defective items in a production cycle, businesses can better manage quality and make informed decisions on improvement areas. In the exercise, we look at the likelihood of finding different quantities of defective screws when selecting **10 screws from the daily production**.

The binomial distribution helps model situations where there are two possible outcomes, like defective or non-defective items. It's ideal for the scenario in the exercise where screws are either **good** or **defective**.

The binomial distribution formula, \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} \] calculates the probability of getting exactly **k defective items** out of **n trials**, where each trial has the probability **p** of being defective. Here, **n = 10** and **p = 0.2**. For example, to find the probability of exactly 2 defective screws (part a), we substitute the known values into the formula: \[ P(X = 2) = \binom{10}{2} (0.2)^2 (0.8)^8 \] This allows us to calculate the likelihood of specific numbers of defective items, helping in understanding and managing quality in production.

Cumulative probability refers to the probability of getting up to a certain number of successes. In part (b) and part (c) of the exercise, we deal with cumulative probabilities. Instead of looking at the probability of exactly **k defective screws**, we find the chance of **more than 2** or **more than 5** defective screws.

For part (b), we need to calculate the probability of having more than 2 defective screws, which is found by subtracting the cumulative probability of **2 or fewer** defective screws from 1: \[ P(X > 2) = 1 - (P(X=0) + P(X=1) + P(X=2)) \]

For part (c), we compute the probability of having more than 5 defective screws by finding the cumulative probability of **5 or fewer**: \[ P(X > 5) = 1 - P(X \text{ ≤ } 5) \] Cumulative probabilities are useful because they help us understand ranges of outcomes and are critical in quality control and risk management.